SchedulingSeminar exercises

2010.10.14Process and Production

Management

Scheduling• Estabilishing the timing of the use of equipment,

facilities and human activities in an organization.– High volume systems (characterized by standardized

equipment and activities that provide identical or highly similar operations on products – autos, personnal computers, radios, TVs)

– Intermediate – volume systems (ouput fall between the standardized type of output of the high volume systems and made-to-order output of job shops. Work centers periodically shift from one product to another, but lot size relatively high.

– Low-volume systems (products are made to order, and orders differ in terms of processing requirements, materials needed, processing time.)

Priority rules• Select the order in which job will be processed• First come first served (FCFS) – jobs are

processed in the order which they arrive at a machine or work center

• Shortest processing time (SPD) – according to processing time, shortest job first

• Earliest due date (EDD) – according to due date, earliest due date first.

• Critical ratio (CR) – smallest ratio of time remaining until due date to processing time remaining.

Assumptions

• The set of jobs is known• Setup time is independent of processing

sequence• Setup time is deterministic• Process time are deterministic• There will be no interruptions in

processing such as machine breakdowns, accidents, or workes illness.

Example 1Job/

productProcessing time (days)

Due date

(days)A 2 7B 8 16C 4 4D 10 17E 5 15F 12 18

• Determine the sequence of jobs/products

• Determine average flow time

• Determine average tardiness

• Determine average number of jobs at the workcenter

• Job flow time – the length of time a job is at a particular workstation or work center. (it includes not only procesing time, but waiting time, transportation time)– Average flow time = total flow time /number of jobs

• Job lateness – length of time the job completion date is exceed the date the job was due to promised to the customer– Average tardiness=total lateness/number of products

• Makespan – is the total time needed to complete a group of jobs (time between start the first and completion of the last one)– Average number of jobs=total flow time/makespan (it

reflects the average work-in-process inventory if the jobs represent equal amount of inventory)

FCFSJob/

productProcessin time (days)

Due date (days)

(2)

Flow time(3)

Days tardy (0 if negative)(3)-(2)

A 2 7 2 2-7=-50B 8 16 2+8=10 10-16=-60C 4 4 10+4=14 14-4=10D 10 17 14+10=24 25-17=7E 5 15 24+5=29 29-15=14F 12 18 29+12=41 41-18=23

41 120 54

• Aft=120/6=20 days• At=54/6=9 days• M=41 Aj=120/41=2,93 pieces

ABCDEF

SPTJob/

productProcessin time (days)

Due date (days)

(2)

Flow time(3)

Days tardy (0 if negative)(3)-(2)

A 2 7 2 2-7=-50C 4 4 2+4=6 6-4=2E 5 15 5+6=11 11-15=-40B 8 16 11+8=19 19-16=3D 10 17 19+10=29 29-17=12F 12 18 29+12=41 41-18=23

41 108 40

• Aft=108/6=18 days• At=40/6=6,67 days• Aj=108/41=2,63 pieces

ACEBDF

EDDJob/

productProcessin time (days)

Due date (days)

(2)

Flow time(3)

Days tardy (0 if negative)(3)-(2)

C 4 4 4 4-4=0A 2 7 4+2=6 6-7=-10E 5 15 5+6=11 11-15=-40B 8 16 11+8=19 19-16=3D 10 17 19+10=29 29-17=12F 12 18 29+12=41 41-18=23

41 110 38

• Aft=110/6=18,33 days• At=38/6=6,33 days• Aj=110/41=2,68 pieces

CAEBDF

CRJob/

productProcessin time (days)

Due date

(days)(2)

CR0

A 2 7 (7-0)/2=3,5

B 8 16 (16-0)/8=2

C 4 4 (4-0)/4=1

D 10 17 (17-0)10=1,7

E 5 15 (15-0)/5=3

F 12 18 (18-0)/12=1,5

41

Remaining time untill due to dat (on the 0th day)

1.

CR4

(7-4)/2=1,5

(16-4)/8=1,5

-------

(17-4)/10=1,3

(15-4)/5=2,2

(18-4)/12=1,172.

???

Job/ product CR16

A (7-16)/2=-4,5

B (16-16)/8=0

C -------

D (17-16)/10=0,1

E (15-16)/5=-0,2

F -----

CR18

--------

(16-18)/8=-0,25

-------

(17-18)/10=-0,1

(15-18)/5=-0,6

---------

1.

2.

3.

4.

CR23

--------

(16-23)/8=-0,875

-------

(17-23)/10=-0,6

-------

---------

4.

5.

CRCFAEBDJob/

productProcessin

time (days)

Due date (days)

(2)C 4 4F 12 18A 2 7E 5 15B 8 16D 10 17

41

Flow time(3)

44+12=1616+2=1818+5=2323+8=3131+10=41133

• Aft=133/6=22,16 days• At=58/6=9,66 days• Aj=133/41=3,24 pieces

Days tardy (0 if negative)(3)-(2)4-4=016-18=-2018-7=1123-15=831-16=1541-17=2458

Sequencing Jobs through two Work Center

• Job time is known and constant• Job time is independent of job sequence• All job must follow the same two-step job

sequence• Job priorities cannot be used• All units must be completed at the first

work center before moving on to the second work center

Example 2Job/

product#1 #2

A 5 5B 4 3C 8 9D 2 7E 6 8F 12 15

• Select the job with the shortest time. – If the shortest time at

first wc, schedule that job first.

– If the time at the 2nd wc, schedule the work last.

• Eliminate the job and its time from further consideration

• Repeat steps

D BAE C F

Chart

D E C F A B#1

D E C F A B#2

2 8

1512F86E72D98C34B55A

#2#1Job/ product

1512F86E72D98C34B55A

#2#1Job/ product

16 28 33 37

2 9 17 26

Idle time

28 43 48 51

Flow time: 51 hours

Thank you for your attention!